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    Maths-
    General
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    Question

    The solution of the differential equation fraction numerator d y over denominator d x end fraction plus y equals cos invisible function application x is

    1. y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent    
    2. y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x minus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent    
    3. y equals cos invisible function application x plus sin invisible function application x plus c e to the power of negative x end exponent    
    4. None of these    

    The correct answer is: y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent

      It is linear equation of the form fraction numerator d y over denominator d x end fraction plus P y equals Q
      So, I.F. equals e to the power of not stretchy integral 1 d x end exponent equals e to the power of x end exponent
      Hence solution is y. e to the power of x end exponent equals not stretchy integral cos invisible function application x. e to the power of x end exponent d x plus c
      Þ y equals fraction numerator 1 over denominator 2 end fraction left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis plus c e to the power of negative x end exponent.

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